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Numerical · Q20

Q.For a certain metal surface, light of frequency 8×1014 Hz8\times10^{14}\ \text{Hz} gives a stopping potential of 0.6 V0.6\ \text{V}. Calculate the work function of the metal (in eV) and the threshold frequency of the metal.

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Photon energy at the given frequency. hν=6.63×10−34×8×1014=5.304×10−19 J≈3.315 eVh\nu=6.63\times10^{-34}\times8\times10^{14}=5.304\times10^{-19}\ \text{J}\approx3.315\ \text{eV}

Work function. From eV0=hν−ϕ0eV_0=h\nu-\phi_0 (with V0=0.6 VV_0=0.6\ \text{V}, so eV0=0.6 eVeV_0=0.6\ \text{eV} numerically): ϕ0=hν−eV0≈3.315−0.6=2.715 eV (≈2.72 eV)\phi_0=h\nu-eV_0\approx3.315-0.6=2.715\ \text{eV}\ (\approx2.72\ \text{eV})

Threshold frequency. Converting ϕ0\phi_0 to joules (2.715×1.6×10−19≈4.34×10−19 J2.715\times1.6\times10^{-19}\approx4.34\times10^{-19}\ \text{J}): ν0=ϕ0h=4.34×10−196.63×10−34≈6.55×1014 Hz\nu_0=\frac{\phi_0}{h}=\frac{4.34\times10^{-19}}{6.63\times10^{-34}}\approx6.55\times10^{14}\ \text{Hz} …

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